Lemma 15.35.15.A9F34BD594E1 For every $\tau >0$, there is a nonnegative $t_\tau \in \mathrm{PSH}(X,h)\cap C(X)$ such that every exponential subsolution at an exponent $\lambda \geq \tau $ lies below $t_\tau $. Put $K_\tau =\sup _ Xt_\tau $. If $\tau \leq \lambda _1\leq \lambda _2$ belong to $\Lambda _ h(f)$, then
\[ U_{\lambda _2}\geq U_{\lambda _1} -K_\tau \frac{\lambda _2-\lambda _1}{\lambda _2}. \]
Consequently, $\lambda \mapsto U_\lambda +K_\tau \log (\lambda /\tau )$ is nondecreasing on $\Lambda _ h(f)\cap [\tau ,+\infty )$.
Proof.
Set $q=\min (f,1)$, a nonzero bounded density. By theorem 14.19.3, there is a continuous $w\in \mathrm{PSH}(X,h)$ with $\mathrm{MA}_ h(w)=\mathrm e^{\tau w}q\, \mathrm dV$. Put $t_\tau =w+\max (0,-\inf _ Xw)$. Then $t_\tau \geq 0$ and $t_\tau \geq w$, so for $\lambda \geq \tau $,
\[ \mathrm{MA}_ h(t_\tau )=\mathrm e^{\tau w}q\, \mathrm dV \leq \mathrm e^{\lambda t_\tau }f\, \mathrm dV. \]
Let $u$ be an exponential subsolution at an exponent $\lambda \geq \tau $, and let $\epsilon >0$. On $\{ t_\tau <u-\epsilon \} $,
\[ \mathrm{MA}_ h(t_\tau ) \leq \mathrm e^{\lambda (u-\epsilon )}f\, \mathrm dV \leq \mathrm e^{-\lambda \epsilon } \mathrm{MA}^{\mathrm{ext}}_ h(u-\epsilon ). \]
The bounded function $t_\tau $ satisfies $P_{h/2}(t_\tau )\not\equiv -\infty $, and $u-\epsilon \preceq t_\tau $. Thus theorem 15.2.2 gives $t_\tau \geq u-\epsilon $, and letting $\epsilon \downarrow 0$ gives $t_\tau \geq u$.
For the order estimate, put $\delta =K_\tau (\lambda _2-\lambda _1)/\lambda _2$. By the first part, $U_{\lambda _1}\leq t_\tau \leq K_\tau $, so $(\lambda _2-\lambda _1)U_{\lambda _1}\leq \lambda _2\delta $, that is, $\lambda _1U_{\lambda _1}\geq \lambda _2(U_{\lambda _1}-\delta )$. Since the Monge–Ampère measure is unchanged by adding constants,
\begin{align*} \mathrm{MA}^{\mathrm{ext}}_ h(U_{\lambda _1}-\delta ) & =\mathrm e^{\lambda _1U_{\lambda _1}}f\, \mathrm dV\\ & \geq \mathrm e^{\lambda _2(U_{\lambda _1}-\delta )}f\, \mathrm dV. \end{align*}
Thus $U_{\lambda _1}-\delta $ is an exponential subsolution at $\lambda _2$, and the maximality of $U_{\lambda _2}$ gives $U_{\lambda _2}\geq U_{\lambda _1}-\delta $. Finally, $1-x^{-1}\leq \log x$ for $x\geq 1$; with $x=\lambda _2/\lambda _1$ this gives $\delta \leq K_\tau \log (\lambda _2/\lambda _1)$, which proves the monotonicity.
$\square$
Pome
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